Given: isosceles trapezoid efgh prove: δfhe ≅ δgeh it is given that trapezoid efgh is an isosceles trapezoid. we know that fe ≅ gh by the definition of . the base angle theorem of isosceles trapezoids verifies that angle is congruent to angle . we also see that eh ≅ eh by the property. therefore, by , we see that δfhe ≅ δgeh.

Given: An Isosceles trapezoid EFGH in which EF =GH

To prove: ΔFHE ≅ ΔGEH

Proof: In Isosceles trapezoid EFGH, Considering two triangles ΔFHE and ΔGEH

  1. FE ≅ G H   → [ Given]

  2. ∠H = ∠E

→ Draw GM⊥HE and FN ⊥EH, and In Δ GMH and ΔFNE,

                          GH=FE [Given]

          ∠M+∠N=180° so GM║FN and GF║EH, So GFMN is a rectangle.]

∴ GM =FN [opposite sides of rectangle]

∠GMH = ∠FNE [ Each being 90°]

Δ GMH ≅ ΔFNE  [ Right hand side congruency]

→∠H =∠E [CPCT]

→ Side EH is common i.e  EH ≅ EH .

→ΔFHE ≅ ΔGEH. [SAS]

ok then what happened.

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier